How would we go about solving the equation

f'(x)=f(-x)?

My thinking:
$f(x)=\int f(-x)\mathrm{dx}$

sub $-b=x$,

$f(-b)=\int -f(b)\mathrm{db}$
So we are given f(- something) in terms of f(something). It would seem as though the b is a dummy variable and could be replaced by a y,z, or an x, so for convenience, we'll replace it by an x.

We are given

$d/\mathrm{dx}(f(x))=\int -f(x)\mathrm{dx}$
at this point, we will replace f(x) with y. we get

d² y/dx² =-y

Giving [as one solution]

dy/dx=iy, so y=e^ix , which doesn't seem to satisfy the initial equation.

Brad

I think the problem is that while d² y/dx² = -y is implied by dy/dx = -y, the converse is not true. This just means that you need to further qualify your answer.
So, from f''(x) = -f(x), you arrive at the solution f(x) = e ^ix = A cos x + B sin x.
So f'(x) = - A sin x + B cos x.
Now f(-x) = A cos -x + B sin -x = A cos x - B sin x.
So if f'(x) = f(-x) then -A sin x + B cos x = A cos x - B sin x and therefore A = B.
So the solution to the equation f'(x) = f(-x) is f(x) = A (cos x + sin x).

Tom.

Thanks. I'd never encountered an equation before that wouldn't give one specific solution if you dropped a constant, but I guess you have to keep the constants in for this example and then further solve.

Thanks again,

Brad

I should have said, the general solution to f''(x) = -f(x) is f(x) = Ce^ix + De^-ix .

Tom.

Tom, I was gonna ask you about that actually.

Originally, you wrote
"you arrive at the solution f(x) = e^ix = Acosx + Bsinx"
That's not true is it?

You then said gen soln should have been f(x) = Ce^ix +De^-ix

Does that explain the mistake? This confused me because I've never seen the general solution to such a differential eqn written in that form (i.e. the exponential form)

thanks

The general solution to f''(x) = -f(x) is
f(x) = Ce^ix + De^-ix = A cos x + B sin x . If you want, I can go into details on how you arive at this solution, but you should be able to check it yourself by finding the second derivative of f(x).
The general solution to differential equations of the form g(x) = af(x) + bf'(x) + cf''(x) + ... will involve the exponential form.

The reason I can then write f(x) = Ce^ix + De^-ix = A cos x + B sin x is because
Ce^ix = C cos x + C.i sin x
De^-ix = D cos -x + D.i sin -x = D cos x + -D.i sin x
The general solution can then be written as
f(x) = (C+D) cos x + (C-D)i sin x
Because we are only interested in real solutions, we then make two definitions, that
C+D = A
(C-D)i = B
where A and B are real. We can do this because C and D can take any value, so for any A, B, we can find some C, D such that this is true. I hope that helps see how I arrived at that solution.

f(x) = e^ix is a specific solution, but not the general solution.

Tom.